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How would I evaluate the integral: $$I=\int{(\cos(x)\cosh(x)+\sin(x)\sinh(x)})\,dx$$ My thought was to use: $$\cos(ix)=\cosh(x)$$ and $$\sin(ix)=i\sinh(x)$$ or expand all four trig functions into exponentials but this was very messy

EDIT:

If I split this into two integrals where $I=I_1+I_2$ $$I_1=\int{cos(x)cosh(x)}dx$$ $$I_2=\int{sin(x)sinh(x)}dx$$

$$I_1=sin(x)cosh(x)-\int{sin(x)sinh(x)}dx$$ This second part is equal to $I_2$ so does that mean that $$I=sin(x)cosh(x)$$

Henry Lee
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4 Answers4

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Hint...try differentiating $\sin x\cosh x$

David Quinn
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    (+1) for the monster hint. – Mark Viola May 23 '18 at 19:14
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    Not a hint as much as just giving them the answer – Dylan May 23 '18 at 19:15
  • My thoughts exactly – mallan May 23 '18 at 19:17
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    I'm actually going to downvote this. – Dylan May 23 '18 at 19:20
  • Someone downvoted my answer. I would be interested to know why. For what it's worth, I am making the observation that the expression to be integrated is an exact derivative which one should be able to recognise. – David Quinn May 23 '18 at 19:21
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    I downvoted. I gave my reasons. There is no hint here, you just told them the answer. There's no learning process. – Dylan May 23 '18 at 19:22
  • @Dylan actually I'm not giving the OP the answer. The OP actually has to follow the hint to find out what the answer is. – David Quinn May 23 '18 at 19:32
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    The problem is your hint is the answer. It does not tell OP how to arrive at the anti-derivative if he didn't know this expression, which he clearly didn't. I'm saying you need to give more subtlety with the "hint" so OP can figure it out for himself. This is too obvious. – Dylan May 23 '18 at 19:35
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    @Dylan: I disagree that there is no learning. Trying to differentiate an ansatz (obtained from educated guess) is a quite acceptable integration technique, which is worth to know. –  May 23 '18 at 19:41
  • @Dylan I think my point is that even though you know what the answer is, the OP does not know what the answer is until he acts on the hint. In fact I'm not sure he yet knows what the answer actually is! – David Quinn May 23 '18 at 19:46
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    @YvesDaoust It would be better to tell them how to make the ansatz, rather than give them the ansatz straight away. How can you apply this knowledge without knowing all the steps? – Dylan May 23 '18 at 19:47
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    @DavidQuinn To me it's like spoiling a movie. Once you gave this hint, it's obvious that it's going to be the answer. In the end he still wouldn't know how to go back and do the integral. All he knows is the answer to the problem.

    Anyway, he followed my hint of IBP and came to the anti-derivative, so that answers your question (of whether he knows the answer now).

     – Dylan May 23 '18 at 19:50
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Your integrand is not exactly in a suitable form. You'd better consider the following two functions:

$$\cos x\cosh x-i\sin x\sinh x=\cos(x+ix)=\cos(1+i)x,$$ and $$\cos x\cosh x+i\sin x\sinh x=\cos(x-ix)=\cos(1-i)x.$$

The antiderivatives are simply

$$\frac{\sin(1+i)x}{1+i},\frac{\sin(1-i)x}{1-i}$$ which you can develop in terms of the trigonometric and hyperbolic functions. Finally, take the sum of the real and imaginary parts. And laugh.

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I think you might want to notice that we have $\sin x$ on right side while it's derivative i.e. $\cos x$ on left side. Also we have $\cosh x$ on left side while it's derivative i.e. $\sinh x$ on the right side.

Don't you think this seems to be much similar to what happens in the product rule.

Let $f(x)=\sin x$ and $g(x)=\cosh x$

Hence the integral can be written as $$\int (f'g +g'f) dx$$

Which simply equals $$f(x)\cdot g(x)+ C$$ by noticing the product rule.

Hence the answer to integral is $$\sin x\cosh x +C$$

You might also want to solve One such question from MIT Integration bee using similar idea which is $$\int (\sin (101x) \cdot \sin^{99}x)dx$$

Rohan Shinde
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    Off-Topic: I found that MIT Integration Bee Integral!! https://youtu.be/qQ-56b_LvOw at 1:17:35. That was amazing! – manooooh May 24 '18 at 02:18
  • $\int{sin(101x).sin^{99}(x)}dx=\int{sin^{99}(x).cos(x).sin(100x)+sin^{100}(x).cos(100x)}dx$ and $\int{sin^{99}(x).cos(x)}dx=\frac{sin^{100}(x)}{100}$ then use the same method as was used before where the second part of the integral $I_2$ is equal to the product of doing IBP on $I_1$ – Henry Lee May 24 '18 at 08:03
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    @HenryLee Or else take $f(x)=\sin^{100} x$ and $g(x)=\sin (100x)$ and continue – Rohan Shinde May 24 '18 at 08:10
  • it is quite a nice integral though, thats why I often do the MIT integration bee integrals, and the original one was from the qualifying test: http://www.mit.edu/~same/pdf/qualifying_round_2018_test.pdf – Henry Lee May 24 '18 at 08:13
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$$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx$$ First divide this into two integrals $$\int \cos(x) \cosh(x)dx+\int \sin(x) \sinh(x) dx$$ Then solve for $\int \cos(x) \cosh(x)dx$ using integration by parts and we get, $$\int \cos(x) \cosh(x)dx=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x))$$ and $$\int \sin(x) \sinh(x)dx=\frac12(\sin(x)\cosh(x)-\cos(x)\sinh(x))$$

Now add them both together and we get, $$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x))+\frac12(\sin(x)\cosh(x)-\cos(x)\sinh(x))$$ $$=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x)+\sin(x)\cosh(x)-\cos(x)\sinh(x))$$ $$=\frac12(2 \sin(x)cosh(x))$$ $$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx=\sin(x)\cosh(x)+C$$

tien lee
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